If the length of a $29 c m$ $29 cm$ spring increases to $47 c m$ $47 cm$ when a $15 k g$ $15 kg$ weight is hanging from it, what is the spring's constant?

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If the length of a $29 c m$ $29 cm$ spring increases to $47 c m$ $47 cm$ when a $15 k g$ $15 kg$ weight is hanging from it, what is the spring's constant?

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Answer 1 · 2016-03-24T14:46:14

$= 833.33 N m^{- 1}$ $=833.33Nm^-1$

Explanation:

$F = k x$ $F=kx$
force $F = m g = 15 \times 10 N = 150 N$ $F=mg=15xx10N=150N$
$g = 10 m s^{- 2}$ $g =10 ms^-2$
Increase in length $l = (47 - 29) c m = 18 c m = 0.18 m$ $l=(47-29)cm=18cm=0.18m$
force constant $k = \frac{F}{x} = \frac{150}{0.18} \frac{N}{m} = 833.33 N m^{- 1}$ $k=F/x=150/0.18N/m=833.33Nm^-1$

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If the length of a $29 c m$ $29 cm$ spring increases to $47 c m$ $47 cm$ when a $15 k g$ $15 kg$ weight is hanging from it, what is the spring's constant?

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Explanation:

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If the length of a 29cm29 cm spring increases to 47cm47 cm when a 15kg15 kg weight is hanging from it, what is the spring's constant?

1 Answer

Explanation:

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If the length of a $29 c m$ $29 cm$ spring increases to $47 c m$ $47 cm$ when a $15 k g$ $15 kg$ weight is hanging from it, what is the spring's constant?